STRUCTURE OF Sm-144, Sm-149, Sm-150, Sm-152, Sm-154
By Prof Lefteris Kaliambos (Natural philosopher in New Energy) ( July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of electromagnetic laws in favor of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for discovering the nuclear force and structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). ' ' STRUCTURES OF SAMARIUM Naturally occurring samarium (Sm) is composed of five stable isotopes, Sm-144, Sm-149, Sm-150, Sm-152 and Sm-154, and two extremely long-lived radioisotopes, Sm-147 (1.06×1011y) and Sm-148 (7×1015y), with Sm-125 being the most abundant (26.75% natural abundance) Comparing the structures of Samarium of 62 protons with the structures of Nd of 60 protons (even number) we see that the stable nuclides of Samarium have the same symmetry as those of Nd. (See my STRUCTURE OF Nd-142....Nd-148 ). After a careful analysis of this comparison I discovered that the p37 and p38 of Nd are replaced here by the p61 and p62. So the two horizontal squares here are not changed. Under this condition the number N of blank positions is given by The two squares give 8n of strong bonds with opposite spins. The first and the sixth plane give 4(n) of weak bonds with opposite spins. The second and the fifth plane give 4{n} with three bonds per neutron and 8n of opposite spins The third and the fourth plane give 4(n) of weak bonds with opposite spins The p61 and p62 give 4(n) of opposite spins. That is N= 4{n} +16n + 12(n) = 32 blank positions able to receive 32 extra neutrons of opposite spins . ' ' DIAGRAM OF SAMARIUM FORMING 32 BLANK POSITIONS Here the additional p61 and p62 are shown near the n47 and n48 respectively for making symmetrical vertical rectangles. Note that the p47n47 along with the p48n48 make two symmetrical alpha particles of opposite spins . But you cannot see the p49n49, the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' n40......p40........n' ' n......... p38.......n38 H. Square with n' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth h. plane' ' n........p29.........n10.........p10…… n30 ' ' n29…… p9..........n9 …….p30.........n Fifth h. plane' ' p47.......n27.........p8..........n8.........p28......... n48....p62' ' n45...........p27........n7.........p7........n28..........p46...........(n ) Fourth h. plane ' ' p61......n47..........p25.........n6.........p6..........n26...........p48' ' (n)....p45..........n25……….p5..........n5……….p26.........n46 Third h. plane' ' n23………p4........n4………….p24..............n' ' n.......p23…….....n3………p3………..n24 Second h. plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22] First h. plane' ' n.........p37........n37 ' ' n39......p39........n H. Square with n' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' Here you see the 2(n) of weak horizontal bonds ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' ' ' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE' Here we have 2{n} +4n of negative spins and the same situation occurs at the fifth horizontal plane with positive spins. ' n' ' n14.......p14........{n}' ' n23.......p4.........n4.........p24..........n ' ' n.......p23........n3........p3.........n24' ' {n}...... p13......n13 ' ' n' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' (n.)........p58....... n50.......p51....n60 ' ' (n) p53........n42........p16......n16......p44.........n54' ' p61 n47........p25........n6........p6........n26.........p48' ' (n) p45........n25........p5........n5........p26........ n46' ' n55........p41.......n15.......p15.......n43...... .p56' ' n57.......p49.......n52...... p59........(n)' Here you can see the following characteristics of the fundamental shapes formed by the nucleons of the central parallelepiped as The p5n5 and n6p6 create the small horizontal square of Mg-24 for creating the central parallelepiped of the alpha particle nuclei. The n15p15 and p16n16 create the first small horizontal rectangle. The p25n25 and p26n26 create the second small horizontal rectangle. The p41, n42, n43 and p44 make the great horizontal square of the great central parallelepiped. The p45, n46, n47 and p48 form the first great horizontal rectangle. The p49, n50, p51 and n52 form the second great horizontal rectangle. Then these great horizontal rectangles which give the structure of Tellurium with 52 protons are able to form 8 blank positions for receiving 8 vertical pn systems. Here you see the p53, n54, n55, p56 n57, p58, p59 and n60 of such vertical pn systems. Also here you see the p61 with 2(n) , while the p62 with 2(n) are not shown, because they are at the fourth horizontal plane. That is the third and the fourth horizontal plane under the contributions of the vertical p61n61 and the vertical n62p62 give extra 4(n). That is the third and the fourth horizontal plane under the contributions of p61 and p62 give the total number of 8(n). In the structure of the Tellurium the n54, n55, n57 and n60 of the third plane along with the symmetrical positions of the fourth horizontal plane were able to receive 8 extra neutrons . Similarly in the structure of Sm using the top view of the third horizontal plane one concludes that under the contributions of p61 and p62 there exist the same 8 blank positions of the third and the fourth plane for receiving 8 extra neutrons of weak binding energy. STRUCTURE OF Sm-144, Sm-150, Sm-152 AND Sm-154 WITH S =0 Since the stable Sm-144 of 20 extra neutrons has S = 0 we conclude that it is due to the extra 20 neutrons of opposite spins. That is, the Sm-144 of S = 0 has 4{n} +16n of opposite spins. Similarly the Sn-150 with 26 extra neutrons has 4{n} +16n +6(n),of opposite spins, while the Sn-152 of 28 extra neutrons has 4{n} +16n + 8(n). of opposite spins. Finally the Sn-154 of 30 extra neutrons has 4{n} + 16n + 10(n) of opposite spins. STRUCTURE OF Sm-149 WITH S = -7/2 In this case the vertical p61n61 makes with the vertical p45n45 a vertical rectangle which is symmetrical to the rectangle made by n62p62. Therefore the p61 along with the p62 exist at the same fourth plane with negative spins . Especially the p61 along with the p62 produce 4 blank positions of negative spins able to receive 4(n) with S = -2. Therefore such a structure of 32 blank positions has 4{n} +16n + 8(n) = 28 extra neutrons of opposite spins in which we add 4(n) of negative spins. Since the Sm-149 of 25 extra neutrons has S =-7/2 we conclude that it has 4(n) of S =-2 and 21 more neutrons of a total spin S = -3/2. Finally we get 16 extra neutrons of negative spins and 9 extra neutrons of positive spins. Especially the Sm-149 has 4(n) +2{n} +8n + 2(n) = 16 extra neutrons of negative spins and 2{n} + 7n = 9 extra neutrons of positive spins. Category:Fundamental physics concepts